Replace Values in a Vector Based on Another Vector

Replace values in a vector based on another vector

Working with factors might be faster:

xf <- as.factor(x)
y[xf]

Note, that levels(xf) gives you a character vector similar to your x.lvl. Thus, for this method to work, elements of y should correspond to appropriate elements of levels(xf).

R - Replace items in a list based on another vector

You can use a lapply() to wrap around your list, and use stringi::stri_replace_all_fixed() to replace the text.

library(stringi)

data_to_change$animal_split <- lapply(data_to_change$animal_split, stri_replace_all_fixed, new_names$V1, new_names$V2, vectorize = F)

data_to_change$animal_split
[[1]]
[1] "doggy"  "cat"    "monkey"

[[2]]
[1] "goldfish"

[[3]]
[1] "mouse"    "doggy"    "bunny"    "squirrel"

Replace given value in vector

Perhaps replace is what you are looking for:

> x = c(3, 2, 1, 0, 4, 0)
> replace(x, x==0, 1)
[1] 3 2 1 1 4 1

Or, if you don't have x (any specific reason why not?):

replace(c(3, 2, 1, 0, 4, 0), c(3, 2, 1, 0, 4, 0)==0, 1)

Many people are familiar with gsub, so you can also try either of the following:

as.numeric(gsub(0, 1, x))
as.numeric(gsub(0, 1, c(3, 2, 1, 0, 4, 0)))

Update

After reading the comments, perhaps with is an option:

with(data.frame(x = c(3, 2, 1, 0, 4, 0)), replace(x, x == 0, 1))

Replacement of column values based on a named vector

You could use col :

df$col1 <- vec[as.character(df$col)]

Or in mutate :

library(dplyr)
df %>% mutate(col1 = vec[as.character(col)])
#   col col1 
#   <int> <chr>
# 1     1 a    
# 2     1 a    
# 3     1 a    
# 4     1 a    
# 5     2 b    
# 6     2 b    
# 7     3 c    
# 8     3 c    
# 9     3 c    
#10     3 c    
#11     3 c    

Replace values in one column based on a vector conditionally matching another column

The first step is to realize that defining ranges of integers will not work. Instead, I'll go with a list of number pairs:

badData <- list(c(296,310), c(330,335), c(350,565))

with the understanding that we want to check each $wavelength to be within any of these three ranges. More ranges are supported.

The second thing we can do is write a function that checks if a vector of values is within one or more pairs of numbers. (In this example, we "know" that it will not be in more than one, but that's not critical.)

within_ranges <- function(x, lims)  {
  Reduce(`|`, lapply(lims, function(lim) lim[1] <= x & x <= lim[2]))
}

To understand what this is doing, let's debug it, call it, and see what's going on.

debugonce(within_ranges)
within_ranges(df$wavelength, badData)
# debugging in: within_ranges(df$wavelength, badData)
# debug at #1: {
#     Reduce(`|`, lapply(lims, function(lim) lim[1] <= x & x <= 
#         lim[2]))
# }

Let's just run that inner portion:

# Browse[2]> 
lapply(lims, function(lim) lim[1] <= x & x <= lim[2])
# [[1]]
#  [1]  TRUE  TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
# [[2]]
#  [1] FALSE FALSE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE FALSE FALSE
# [[3]]
#  [1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE  TRUE  TRUE

So the first element (T,T,F,F,...) is whether the values (x) fall within the first number pair (296 to 310); the second element with the second pair (330 to 335); etc.

The Reduce( part calls the first argument, a function, on the first two arguments, saves the return, and then runs the same function on the return and the third argument. It stores it, then runs the same function on the return and fourth argument (if exists). It repeats this along the entire length of the provided list.

In this example, the function is the literal | (escaped since it is special), so it is "OR"ing the [[1]] vector with the [[2]] vector. You can actually see what is happening if you add accumulate=TRUE:

# Browse[2]> 
Reduce(`|`, lapply(lims, function(lim) lim[1] <= x & x <= lim[2]), accumulate=TRUE)
# [[1]]
#  [1]  TRUE  TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
# [[2]]
#  [1]  TRUE  TRUE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE FALSE FALSE
# [[3]]
#  [1]  TRUE  TRUE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE  TRUE  TRUE

The first return is the first vector, unmodified. The second element is the original [[2]] vector ORed with the previous return which is this [[1]] vector (which is the same as the original [[1]]). The third element is the original [[3]] vector ORed with the previous return, which is this [[2]]. This results in the three groupings of TRUE (1, 2, 7, 11, 12) that you are expecting. So we want the [[3]] element, which is what we get without accumulating:

# Browse[2]> 
Reduce(`|`, lapply(lims, function(lim) lim[1] <= x & x <= lim[2]))
#  [1]  TRUE  TRUE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE  TRUE  TRUE

Okay, so let's Quit out of the debugger, and give it a full go:

within_ranges(df$wavelength, badData)
#  [1]  TRUE  TRUE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE  TRUE  TRUE

This output looks familiar.

(BTW: inside our function, we could also have used

rowSums(sapply(lims, ...)) > 0

and it would have worked just as well. For that, though, you need to realize that sapply should be returning a matrix with as many columns as we have rows of data in df, odd if you aren't familiar.)

Now, we can NAify what we need to either with dplyr:

df %>%
  mutate(
    reflectance = if_else(within_ranges(wavelength, badData), NA_real_, reflectance)
  )
#    wavelength reflectance
# 1    300.0000          NA
# 2    305.0087          NA
# 3    310.0173   -11.01733
# 4    315.0260   -16.02600
# 5    320.0347   -21.03467
# 6    325.0433   -26.04333
# 7    330.0520          NA
# 8    335.0607   -36.06067
# 9    340.0693   -41.06934
# 10   345.0780   -46.07800
# 11   350.0867          NA
# 12   355.0953          NA

Edit: or another dplyr, using your first thought of replace (not my first by habit, no reason):

df %>%
  mutate(
    reflectance = replace(reflectance, within_ranges(wavelength, badData), NA_real_)
  )

or base R:

df$reflectance <- ifelse(within_ranges(df$wavelength, badData), NA_real_, df$reflectance)
df
#    wavelength reflectance
# 1    300.0000          NA
# 2    305.0087          NA
# 3    310.0173   -11.01733
# 4    315.0260   -16.02600
# 5    320.0347   -21.03467
# 6    325.0433   -26.04333
# 7    330.0520          NA
# 8    335.0607   -36.06067
# 9    340.0693   -41.06934
# 10   345.0780   -46.07800
# 11   350.0867          NA
# 12   355.0953          NA

---

Notes:

• I'm specifically using NA_real_, both for clarity (did you know there are different types of NA?), and partly because in the use of dplyr::if_else, it will complain/fail if the classes of the "true" and "false" arguments are not the same (NA is technically logical, not numeric as your reflectance is);
• I use dplyr::if_else for the first example, since you're already using dplyr, but in case you choose to forego dplyr (or somebody else does), then the base-R ifelse works, too. (It has its liabilities, but it appears to work just fine here.)

Replacing vector elements based on indices of another vector

I guess you want this indexing

> b[a]
[1] 0.5 2.0 3.0 2.5 2.5 1.0 2.0 3.0 0.5

In R how would you replace values in a matrix that have a certain condition with values from another vector?

Thought you had to use apply(Matrix, \(x) pmin(x, Vector)), but actually, you can just use pmin() directly on your Matrix because it will recycle the Vector to match the length.

pmin(Matrix, Vector)
#>      [,1] [,2]
#> [1,]    2    2
#> [2,]    3    3
#> [3,]    3    2
#> [4,]    1    1

---

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